[mark1950]

Calculate the concentration, in mol dm^-3, of hydrated ethanedioic acid in solution KA1 based on the text given.

KA 1 is a solution containing hydrated ethanedioic acid, H₂C₂O₄.2H₂O, and sodium ethanedioate. 25.0 cm³ of KA 1 solution is pipetted into a titration flask and is titrated with KA 3. It is found that 17.20 cm³ of KA 3 is required for a complete reaction. KA 3 contains 1.7 g of hydroxyl ions per dm³.

[DrCMS]

Start with a reaction equation.

How many moles of OH^- react with 1 mole of ethanedioic acid?

Calculate how many moles of hydroxyl ions are in the 17.2 mls of alkali solution?

Using the last two answers now calculate how many moles of ethanedioic acid were in the 25ml of acid solution?

If that are X moles per 25ml how many moles per litre?

[mark1950]

Thanks. I understood all your explanation but I'm stuck at making an equation. There seems to be two chemicals in one solution and I don't seem to be able to figure out the equation. Care for some help? Thanks.

[DrCMS]

Acid + base = ?

[mark1950]

Hm. So you mean that its H₂C₂O₄.2H₂O + Na₂C₂O₄ = H₂O + Na₂C₂O₄? But its impossible to balance this equaion. How about the OH ions that was supplied via titration? How do they come in? Forgive me but I've never come across equations with 3 different reactants before so this whole thing is quite new to me.

[mark1950]

OH wait. I think I got it. Its 2NaOH + H₂C₂O₄.2H₂O = Na₂C₂O₄ + 4H₂O, right? THanks.